Prove: If $G$ Is Abelian Then For The Homomorphism $f:G\to H$, ${\rm im}(f)$ Is Abelian Too

Question

Let $G$ be an abelian group, and let $f:G\to H$ be an homomorphism

Prove: ${\rm im}(f)$ is abelian

Proof: Let $a,b\in G$ then $ab=ba$, lets apply $f$ on both sides: $$f(a)f(b)=f(ab)=f(ba)=f(b)f(a)$$

As $f$ is an homomorphism

The opposite is false, we can take $f:GL_2(\mathbb{R})\to R^*$ be $f(A)=\det(a)$ Let $$g_1=\begin{pmatrix} 1 & 1 \\ 1 & 0 \\ \end{pmatrix},g_2=\begin{pmatrix} 1 & 0 \\ 1 & 1 \\ \end{pmatrix}$$

Then $\det(g_1)\det(g_2)=1\cdot1=\det(g_2)\det(g_1)$

but $g_1\cdot g_2\neq g_2\cdot g_1$

Is it correct?

Answer 1

Yes, this looks correct. Your counterexample works too, and actually its quite good to be included (even though the question does not asks for one explicitly)!